Confusion while proving logdet function to be concave

Question

I was going through this lecture related to convex optimization. It was proved that logdet function is concave. However, I didn't get the derivation at a part

I didn't get how the step marked in red in the given picture was derived. Which property was used

Answer 1

There are three properties being used here:

• the determinant of a product of square matrices of equal size is equal to the product of the determinants of its square, equally sized factors
• the logarithm of a product is equal to the sum of the logarithms of its factors
• the determinant of a matrix is equal to the product of its eigenvalues.

Let's start from

\begin{align} g(t) = \log \det (Z^{1/2}(I + tZ^{-1/2}VZ^{-1/2})Z^{1/2}). \end{align}

The determinant product property yields

\begin{align} g(t) = \log\Big(\det(Z^{1/2})\det(I + tZ^{-1/2}VZ^{-1/2})\det(Z^{1/2})\Big), \end{align}

and by commutativity of multiplication, we can rearrange it so that

\begin{align} g(t) &= \log\Big(\det(Z^{1/2})\det(Z^{1/2})\det(I + tZ^{-1/2}VZ^{-1/2})\Big), \\ &= \log\Big(\det(I + tZ^{-1/2}VZ^{-1/2})\det(Z)\Big). \end{align}

Using the product-sum property of logarithms, this expression becomes

\begin{align} g(t) &= \log\det(I + tZ^{-1/2}VZ^{-1/2}) + \log\det(Z), \end{align}

and we're most of the way there with the $\det(Z)$ term already separated out.

If the eigenvalues of $Z^{-1/2}VZ^{-1/2}$ are $\lambda_{i}$, then:

• the eigenvalues of $tZ^{-1/2}VZ^{-1/2}$ are $t\lambda_{i}$, because the eigenvectors of each matrix are the same
• the eigenvalues of $I + tZ^{-1/2}VZ^{-1/2}$ are $(1 + t\lambda_{i})$ because the eigenvectors of $Z^{-1/2}VZ^{-1/2}$ will also be eigenvectors for $I$.

This aside about eigenvalues is important, because to complete the derivation, we need to use the property that the determinant of a matrix is the product of its eigenvalues:

\begin{align} g(t) &= \log\left(\prod_{i=1}^{n}(1+t\lambda_{i})\right) + \log\det(Z). \end{align}

To complete the derivation, we use the product-sum property of logarithms (again), and get

\begin{align} g(t) = \sum_{i=1}^{n} \log(1 + t\lambda_{i}) + \log\det(Z). \end{align}